Evaluate using Wallis’ Formula.

Question

1. ∫₀^π/2 sin⁴y cos⁷ ydy
2. ∫¹₀ (1-x²) ^5/2 dx

Answer 1

1. ∫₀^π/2 sin⁴y cos⁷ ydy = ∫₀^π/2sin⁴y(1 - sin²y)³d(siny) = ∫₀^π/2(sin⁴y - 3sin⁶y + 3sin⁸y - sin¹⁰y)d(siny) =

= 1/5 - 3/7 + 3/9 - 1/11 = 16/1155

2.∫¹₀ (1-x²) ^5/2 dx = ∫₀^π/2cos⁵θcosθdθ =

x = sinθ; dx = cosθdθ ∫₀^π/2cos⁶θdθ = 5π/32...

x = 0; θ = 0;

x = 1; θ = π/2

Answer 2

The second one is readily converted to a Wallis integral with the substitution x = sin(y)

which results in W₆ = 5 pi/32

The first one can be rearranged as

∫ cos [ sin⁴ - 3 sin⁶ + 3 sin⁸ - sin¹⁰ ]

which is easy since the derivative of sin is cos

The result is 1/5 - 3/7 + 1/3 - 1//1 = 16/1155